An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If two marbles are picked up at random, what is the probability that either both are green or both are yellow ?
Correct Answer: $$\frac{4}{105}$$
Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 2 marbles such that either both are green or both are yellow.
Then,
n (E) = $$\left( {{}^2\mathop C\nolimits_1 + {}^3\mathop C\nolimits_2 } \right)$$ $$ = \left( {1 + {}^3\mathop C\nolimits_1 } \right)$$ = (1 + 3) = 4
And,n (S) = $${}^{15}\mathop C\nolimits_2 = $$ $$\frac{{15 \times 14}}{{2 \times 1}}$$ = 105
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{105}}$$
Let E be the event of drawing 2 marbles such that either both are green or both are yellow.
Then,
n (E) = $$\left( {{}^2\mathop C\nolimits_1 + {}^3\mathop C\nolimits_2 } \right)$$ $$ = \left( {1 + {}^3\mathop C\nolimits_1 } \right)$$ = (1 + 3) = 4
And,n (S) = $${}^{15}\mathop C\nolimits_2 = $$ $$\frac{{15 \times 14}}{{2 \times 1}}$$ = 105
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{105}}$$